AMC 8 · 1999 · #15
Grade 5 countingPick an answer.
AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Only a handful of distributions exist for 2 extra letters across 3 sets, so Tool #2 (Make a Systematic List) sweeps them all without missing any: 3 ways to put both letters in one set plus 3 ways to put one letter in each of two sets — exactly 6 cases. Tool #6 (Guess and Check) is the natural companion: compute the new product for each case and pick the largest. The size of the case list is small enough that listing beats algebra, which keeps the solution accessible.
Multiply the three set sizes: 60 plates is the baseline we subtract at the end.
Three independent choices multiply — the Grade 5 multiplication-as-counting move.
5.NBT.B.5Make A Systematic ListList every way to spread 2 letters: both in one set (3 ways) or one each in two sets (3 ways) — 6 cases.
When the case count is small, writing the full list is the safest way not to miss the winner.
4.OA.A.3Make A Systematic ListMultiply the three updated sizes in each case to get its new plate total.
One multiplication per row turns the case list into a column of totals you can scan.
5.NBT.B.5Guess And CheckThe best new total is 100 (from 5×5×4); subtract the original 60 for the ADDITIONAL plates.
"Additional" means the gain over the starting count — subtract the old total from the best new one.
4.OA.A.3Guess And CheckOnly 6 ways exist to spread 2 new letters across 3 sets — list them, multiply, and the best one is 5 × 5 × 4 = 100, which is 40 more plates than the original 60.
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